Perhaps you might consider adding a series resistor in the Red branch of the attached schematic, and you might also consider adding a ferrite bead in the Blue branch.

The resistor reduces the magnitude of the current peaks through the diodes, reducing the dV/dt slope of the zig-up piece of the ripple waveform. It also forms a lowpass filter (with C1 + C2 + Rdiode) which reduces the amount of high frequency noise passed from mains to IC1 input. The voltage dropped across this resistor eats into the voltage headroom ("dropout voltage") of the regulator, so carefully choose its resistance based on minimum mains voltage, maximum load current, maximum output voltage, and maximum regulator "dropout" specification. And I suggest you choose its wattage to be at least 3x higher than its theoretical power dissipation, both for long term reliability and also to avoid burning your finger when servicing.

The ferrite bead acts as a (lossy) resistor at high frequencies, further reducing RF noise conducted from mains into IC1.

The resistor reduces the magnitude of the current peaks through the diodes, reducing the dV/dt slope of the zig-up piece of the ripple waveform. It also forms a lowpass filter (with C1 + C2 + Rdiode) which reduces the amount of high frequency noise passed from mains to IC1 input. The voltage dropped across this resistor eats into the voltage headroom ("dropout voltage") of the regulator, so carefully choose its resistance based on minimum mains voltage, maximum load current, maximum output voltage, and maximum regulator "dropout" specification. And I suggest you choose its wattage to be at least 3x higher than its theoretical power dissipation, both for long term reliability and also to avoid burning your finger when servicing.

OK. Can you help me calculate it? The transformer outputs 6V when loaded and about 8V with no load. Current is limited to 500mA because the is the maximum the regulator will do. The regulator has a Low Dropout Voltage, 300mV. Maybe 2-4 ohms at 1w?

Quote:

Originally Posted by Mark Johnson

The ferrite bead acts as a (lossy) resistor at high frequencies, further reducing RF noise conducted from mains into IC1.

I was going to use an coil at the blue arrow, this one. Since a coil and a ferrite bead are just both inductors, does one preclude the other? Can you use both? Would this one be suitable?

The transformer outputs 6V when loaded and about 8V with no load. Current is limited to 500mA because the is the maximum the regulator will do. The regulator has a Low Dropout Voltage, 300mV. Maybe 2-4 ohms at 1w?

Dissipation rating = (0.5A x 0.5A x 0.68R) x 3 = 0.51 watts. Me, I'd go ahead and use a 1W resistor. Just to provide a visual reminder on the PCB: "Hey, this resistor is high wattage so it might be hot! Be careful with it."

I recommend you read the Wikipedia article on ferrite beads; a ferrite bead is not "just an inductor". Rather, ferrite beads are lossy circuit elements at high frequencies; they dissipate power. Inductors, being lossless reactances, do not.

I don´t know how important electrolytic ESR is in this circuit but you could also consider the use of Panasonic FM or FR types for C1 (approx. half the ESR compared with FC types). Endurance for FR types is 2000 h to 10000 h at
+105 deg. C.